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How do you prove a number irrational? - Answers
A irrational number is a real number that cant be expressed as a/b where a and b are integers. Or in more simple terms they cannot be written as decimals they just keep on going like Pi. So how can we tell if a number is irrational? Surely we can just check, every digit. Sadly this wont work as numbers just keep going to infinity. So we use the proof by contradiction to do this. Take √2 for example let us make the supposition that √2 is rational then √2 = m/n where m and n are integers with no common factors if we rearrange that equation for a we get a2 = 2b2 2 times anything is even, hence 2b2 is even, and a2 is even a then must be even as if a were odd a2 would also be odd a = 2k where k is an integer 4k2 = 2b2 2k2 = b2 2 times anything is even, hence 2k2 is even, and b2 is even b then must be even as if a were odd b2 would also be odd b = 2m where m is an integer so; 16m2 = 4k2 so there is a common factor and the supposition is incorrect hence √2 is irrational
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How do you prove a number irrational? - Answers
A irrational number is a real number that cant be expressed as a/b where a and b are integers. Or in more simple terms they cannot be written as decimals they just keep on going like Pi. So how can we tell if a number is irrational? Surely we can just check, every digit. Sadly this wont work as numbers just keep going to infinity. So we use the proof by contradiction to do this. Take √2 for example let us make the supposition that √2 is rational then √2 = m/n where m and n are integers with no common factors if we rearrange that equation for a we get a2 = 2b2 2 times anything is even, hence 2b2 is even, and a2 is even a then must be even as if a were odd a2 would also be odd a = 2k where k is an integer 4k2 = 2b2 2k2 = b2 2 times anything is even, hence 2k2 is even, and b2 is even b then must be even as if a were odd b2 would also be odd b = 2m where m is an integer so; 16m2 = 4k2 so there is a common factor and the supposition is incorrect hence √2 is irrational
DuckDuckGo
How do you prove a number irrational? - Answers
A irrational number is a real number that cant be expressed as a/b where a and b are integers. Or in more simple terms they cannot be written as decimals they just keep on going like Pi. So how can we tell if a number is irrational? Surely we can just check, every digit. Sadly this wont work as numbers just keep going to infinity. So we use the proof by contradiction to do this. Take √2 for example let us make the supposition that √2 is rational then √2 = m/n where m and n are integers with no common factors if we rearrange that equation for a we get a2 = 2b2 2 times anything is even, hence 2b2 is even, and a2 is even a then must be even as if a were odd a2 would also be odd a = 2k where k is an integer 4k2 = 2b2 2k2 = b2 2 times anything is even, hence 2k2 is even, and b2 is even b then must be even as if a were odd b2 would also be odd b = 2m where m is an integer so; 16m2 = 4k2 so there is a common factor and the supposition is incorrect hence √2 is irrational
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- og:descriptionA irrational number is a real number that cant be expressed as a/b where a and b are integers. Or in more simple terms they cannot be written as decimals they just keep on going like Pi. So how can we tell if a number is irrational? Surely we can just check, every digit. Sadly this wont work as numbers just keep going to infinity. So we use the proof by contradiction to do this. Take √2 for example let us make the supposition that √2 is rational then √2 = m/n where m and n are integers with no common factors if we rearrange that equation for a we get a2 = 2b2 2 times anything is even, hence 2b2 is even, and a2 is even a then must be even as if a were odd a2 would also be odd a = 2k where k is an integer 4k2 = 2b2 2k2 = b2 2 times anything is even, hence 2k2 is even, and b2 is even b then must be even as if a were odd b2 would also be odd b = 2m where m is an integer so; 16m2 = 4k2 so there is a common factor and the supposition is incorrect hence √2 is irrational
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